110c-lecture13 - Chapter 24: Liquid-Liquid Solutions (pp...

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Chapter 24: Liquid-Liquid Solutions (pp 963-970) Problems: 13, 18, 22, 25, 32, 34, 45, 50 , 51 In Chapter 23, chemical potential ( µ ) described phase equilibria of a pure substance. liq gas µ liq = µ gas (equil) µ liq > µ gas (vaporization) -dn liq = dn gas Chemical potential can be applied to liquid solutions with two volatile components. G = µ 1 n 1 + µ 2 n 2 g l dn dn 1 1 = g l dn dn 2 2 = n V V liq liq O H = 2 T P s s n G G , 1 ln 1 ln 1 = = δ µ T P n V V , 1 1 = T P n P P , 1 1 = n S S liq liq O H = 2 n H H liq liq O H = 2 2 2 1 1 V n V n V + = 2 2 1 1 P n P n P + = n G G n G liq O H liq O H P T liq O H liq O H 2 2 , 2 2 = = =
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Partial Molar Quantities If () 2 1 , , , n n P T Y Y = is a state function (U,H,S,A,G,. ..), its partial molar quantity is: () j i n P T j j j n Y n n P T Y Y = = , , 2 1 , , , eqn 24.4 p964 Partial molar quantities measure how Y changes (dU,dH,dS,dA,dG,. ..) vs n j : For example, .. , 2 2 .. , 1 1 P T P T n V V and n V V = = δ Don’t confuse molar volume of a pure substance: n V V liq O H = 2 or with partial molar volume of a multicomponent solution: ... , , ni P T j j n V V = or is contribution to Y by Component 1 n G G liq O H = 2 = 1 ln 1 n G G s 1 1 Y n is contribution to V by Component 1 1 1 V n 2 2 1 1 V n V n V + = = 1 ln 1 n G s µ Total volume of 2-component solution:
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Partial Molar Volume Consider the volume of a 2-component system: () ... , , , 2 1 n n P T V V = 2 2 1 1 2 , 1 , 2 , 1 , dn V dn V dP P V dT T V dV n n T n n P + + + = At const T, P : j j j j j j i n P T j dn V dn n V dV = = , , 2 2 1 1 n V n V V + = (2-component system) 1.0 0.8 0.6 0.4 0.2 0.0 Mol fraction of water (x 2 ) (x 1 ) What is the total volume of a solution that contains 100 mL H 2 O + 100 mL propanol? () () mL mL mL V mol mL mol g mL g mL mol mL mol g mL g mL V MW density vol n V n V n n V V i j j j 196 69 . 99 21 . 96 / 18 / 02 . 18 / 998 . 0 100 / 72 / 09 . 60 / 803 . 0 100 * 2 2 1 1 = + = + = = + = = NOT 200 mL Propanol: mol mL V / 83 . 74 0 1 = vs. 72 mL/mol for 50:50 mixture Pure water: mol mL V / 06 . 18 0 2 = vs 18 mL/mol for 50:50 mixture
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Gibbs Free Energy (G) for Multicomponent System () ... , , , 2 1 n n P T G G = + + = j j j i n P T j i n T i n P dn n G dP P G dT T G dG , , , ,
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This note was uploaded on 06/22/2010 for the course CHEM 21360 taught by Professor Ame during the Spring '09 term at East Los Angeles College.

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110c-lecture13 - Chapter 24: Liquid-Liquid Solutions (pp...

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